The total energy of a particle executing $SHM$ at an amplitude of $4 \, cm$ is $20 \, J$. What will be its total energy at a displacement of $x = 2 \, cm$ (in $J$)?

If $T$ is the period of $SHM$,then write the period of kinetic and potential energy.

The potential energy function for a particle executing linear simple harmonic motion is given by $V(x) = k x^{2} / 2$,where $k$ is the force constant of the oscillator. For $k = 0.5 \; N m^{-1}$,the graph of $V(x)$ versus $x$ is shown in the figure. Show that a particle of total energy $1 \; J$ moving under this potential must 'turn back' when it reaches $x = \pm 2 \; m$.

The potential energy of a simple harmonic oscillator of mass $2\, kg$ in its mean position is $5\, J.$ If its total energy is $9\, J$ and its amplitude is $0.01\, m,$ its time period would be

The potential energy of a particle executing $S.H.M.$ is $2.5 \ J$,when its displacement is half of its amplitude. The total energy of the particle is .... $J$.

$A$ particle is executing $S.H.M.$ with total mechanical energy $90 \,J$ and amplitude $6 \,cm$. If its energy is somehow decreased to $40 \,J$,then its amplitude will become ........ $cm$.